Finding numbers who digits ^ (digit position) == original number i.e. (89 = 8^1 + 9^2), but the code won't return anythi

Content Index :

Finding numbers who digits ^ (digit position) == original number i.e. (89 = 8^1 + 9^2), but the code won't return anythi
Finding if two numbers have the same digit and then remove them from in the original number in Haskell
Code for finding the greatest product of five consecutive digits in the 1000-digit number
how to get number of possible 4 digit numbers with constraints on possible digits and position of digits
Find the a 4 digit number who's square is 8 digits AND last 4 digits are the original number
Code for finding largest product of 13 consecutive digits of a 1000 digit number does not give required output

Finding numbers who digits ^ (digit position) == original number i.e. (89 = 8^1 + 9^2), but the code won't return anythi

Tag : python , By : Adam Hill

Date : December 05 2020, 12:18 PM

will be helpful for those in need I think, your code doesn't work, because the missing solution 1676 contains two times the 6. So str(i).index(x) will return the wrong index for the second 6 which makes the calculation faulty.
I guess, this is one of the times, where I'd start without list comprehensions and first get the code to work. Is this generally the result you want?

def sum_dig_pow(a, b): 
    result = []
    for number in range(a, b+1):
        calculation = 0
        for position, digit in enumerate(map(int, str(number))):
            calculation += digit**(position + 1)
        if calculation == number:
            result += [number]
    return result

print(sum_dig_pow(1, 2000))

[1, 2, 3, 4, 5, 6, 7, 8, 9, 89, 135, 175, 518, 598, 1306, 1676]

def sum_dig_pow(a, b): 
    result = []
    for number in range(a, b+1):
        calculation = sum([digit**(position + 1) for position, digit in enumerate(map(int, str(number)))])
        if calculation == number:
            result += [number]
    return result

def sum_dig_pow(a, b): 
    return [number for number in range(a, b+1) if sum([digit**(position + 1) for position, digit in enumerate(map(int, str(number)))]) == number]

sum_dig_pow = lambda a, b: [number for number in range(a, b+1) if sum([digit**(position + 1) for position, digit in enumerate(map(int, str(number)))]) == number]

Comments

No Comments Right Now !

Boards Message :

You Must Login Or Sign Up to Add Your Comments .

Share :

Tags :

Finding numbers who digits ^ (digit position) == original number i.e. (89 = 8^1 + 9^2) but the code wont return anythi , how to get number of possible 4 digit numbers with constraints on possible digits and position of digits , Finding if two numbers have the same digit and then remove them from in the ,

Finding if two numbers have the same digit and then remove them from in the original number in Haskell

Tag : haskell , By : alchemist

Date : March 29 2020, 07:55 AM

Does that help Let x and y be the two numbers. Then one can remove the digits in x which it has in common with y like this:

Prelude> import Data.List
Prelude Data.List> let x = 68
Prelude Data.List> let y = 76
Prelude Data.List> read (show x \\ show y) :: Int
8

s <- nub $ intersect (show x) (show y)

map (read . delete s . show) [x, y]

Code for finding the greatest product of five consecutive digits in the 1000-digit number

Tag : java , By : micaleel

Date : March 29 2020, 07:55 AM

will be helpful for those in need charAt() reports 48 for '0' and 49 for '1' etc. as it is the character values.
Subtract 48 from each and try again.

how to get number of possible 4 digit numbers with constraints on possible digits and position of digits

Tag : math , By : user123585

Date : March 29 2020, 07:55 AM

this one helps. So our hint is: n1n2n3n4, use all of 1234 exactly once.
1) There are three places we can put 1 in, leaving us with _1n3n4, _n21n4 and _n2n31.

Find the a 4 digit number who's square is 8 digits AND last 4 digits are the original number

Tag : python , By : Alex S

Date : March 29 2020, 07:55 AM

I wish this helpful for you From the comments on my answer here, the question was asked (paraphrase): , Here is a 1-liner solution without any modules:

>>> next((x for x in range(1000, 10000) if str(x*x)[-4:] == str(x)), None)
9376

>>> next((x for x in range(3163, 10000) if str(x*x)[-4:] == str(x)), None)
9376

Code for finding largest product of 13 consecutive digits of a 1000 digit number does not give required output

Tag : python , By : kiirpi

Date : March 29 2020, 07:55 AM

I think the issue was by ths following , The code is to find the largest possible product of 13 consecutive digits of a 1000 digit number. When I tried to run it on IDLE, it just gave RESTART and the directory where I saved the .py file. When I tried this on Pycharm(I know its not the IDE's problem, but I just had to try), no output. Am I doing something wrong? , Your original code can be simplified to the following.



n=7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450

def product(j):
    p = 1
    for i in range(j,j+13):
        p=p*s[i]
    return(p)
s=[]
for i in range(0,1000):
    s=s+[n%10]
    n=n//10
k = 0
N = len(s)
j = 0
while j<N-13:  #need to 13 elements before end
    c = product(j)
    if c>k:
        k=c
    j = j + 1
print(k)
n=7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450

def product(j):
    p = 1
    for i in range(j,j+13):
        p=p*s[i]
    return(p)
s=[]
for i in range(0,1000):
    s=s+[n%10]
    n=n//10
k = 0
N = len(s)
j = 0
while j<N-13:  #need to 13 elements before end
    c = product(j)
    if c>k:
        k=c
    j = j + 1
print(k)


digits = str(n) # generates digits of number
products = [product(j) for j in range(1000-13)] # list comprehension
                                                # uses you product function
print(max(products))
digits = str(n) # generates digits of number
products = [product(j) for j in range(1000-13)] # list comprehension
                                                # uses you product function
print(max(products))